This repository has been archived by the owner on Aug 6, 2022. It is now read-only.
-
Notifications
You must be signed in to change notification settings - Fork 172
/
ch2exercises.xml
539 lines (354 loc) · 15.9 KB
/
ch2exercises.xml
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
<?xml version="1.0" encoding="UTF-8"?>
<!--********************************************************************
Copyright 2017 Georgia Institute of Technology
Permission is granted to copy, distribute and/or modify this document
under the terms of the GNU Free Documentation License, Version 1.3 or
any later version published by the Free Software Foundation. A copy of
the license is included in gfdl.xml.
*********************************************************************-->
<section xml:id="ch2exercises">
<title>Exercises</title>
<subsection xml:id="subsection-Rn">
<title>Exercises for Chapter 2</title>
<p>
Questions:
<ol>
<li>
State whether the following are true or false. If false, give a counterexample/reason, and if true, explain why it is true.
<ol>
<li>
A system of 5 linear equations can have 5 solutions.
</li>
<li>
The solution set to any consistent system of linear equations is a subspace.
</li>
<li>
The solution set to a consistent system of linear equations of the form <m>Ax=0</m> is a subspace.
</li>
<li>
The intersection of any two subspaces (i.e., the collection of points that are in both subspaces) is a subspace.
</li>
<li>
The union of any two subspaces (i.e., the collection of points that are in at least on of the subspaces) is a subspace.
</li>
</ol>
</li>
<li>
Use Gaussian elimination to find the unique solution for each system.
<ol>
<li>
<me>\begin{aligned}
2x &+ &3y &= &13 \\
x &- &y &= &-1
\end{aligned}
</me>
</li>
<li>
<me>\begin{aligned}
x & & &- &z &= &0 \\
3x &+ &y & & &= &1 \\
-x &+ &y &+ &z &= &4
\end{aligned}</me>
</li>
</ol>
</li>
<li>
Each system of linear equations is in echelon form (that is, the corresponding matrix is in echelon form).
For each, say whether the system has a unique solution,
no solution, or infinitely many solutions.
<ol>
<li>
<me>
\begin{aligned}
-3x &+ &2y &= &0 \\
& &-2y &= &0
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
x &+ &y & & &= &4 \\
& &y &- &z &= &0
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
x &+ &y & & &= &4 \\
& &y & &-z &= &0 \\
& & & &0 &= &0
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
x &+ &y &= &4 \\
& &0 &= &4
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
3x &+ &6y &+ &z &= &-0.5 \\
& & & &-z &= &2.5 \\
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
x &- &3y &= &2 \\
& &0 &= &0
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
2x &+ &2y &= &4 \\
& &y &= &1 \\
& &0 &= &4
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
2x &+ &y &= &0
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
x &- &y &= &-1 \\
& &0 &= &0 \\
& &0 &= &4
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
x &+ &y &- &3z &= &-1 \\
& &y &- &z &= &2 \\
& & & &z &= &0 \\
& & & &0 &= &0
\end{aligned}
</me>
</li>
</ol>
</li>
<li>
Use Gaussian elimination to solve each system
or conclude `many solutions' or `no solutions'.
<ol>
<li><me>
\begin{aligned}
2x &+ &2y &= &5 \\
x &- &4y &= &0
\end{aligned}</me>
</li>
<li><me>
\begin{aligned}
-x &+ &y &= &1 \\
x &+ &y &= &2
\end{aligned}
</me>
</li>
<li><me>
\begin{aligned}
x &- &3y &+ &z &= &1 \\
x &+ &y &+ &2z &= &14
\end{aligned}
</me>
</li>
<li><me>
\begin{aligned}
-x &- &y &= &1 \\
-3x &- &3y &= &2
\end{aligned}
</me>
</li>
<li><me>
\begin{aligned}
& &4y &+ &z &= &20 \\
2x &- &2y &+ &z &= &0 \\
x & & &+ &z &= &5 \\
x &+ &y &- &z &= &10
\end{aligned}
</me>
</li>
<li><me>\begin{aligned}
2x & & &+ &z &+ &w &= &5 \\
& &y & & &- &w &= &-1 \\
3x & & &- &z &- &w &= &0 \\
4x &+ &y &+ &2z &+ &w &= &9
\end{aligned}
</me>
</li>
</ol>
</li>
</ol>
</p>
<p>
Answers:
<ol>
<li>
<ol>
<li>
False. The solution set to any system of linear equations is either empty (inconsistent equations), has a unique solution (consistent equations, no free variables), or infinite (if there is a free variable, since this free variable can then take on the value of any of the infinitely many real numbers).
</li>
<li>
False. For example, the solution set of the set of equations <m>x+y=3</m> in <m>\mathbb R^2</m> is a line, but doesn't pass through the origin.
</li>
<li>
True. Here, we can check the axioms in the definition of a subspace. Given the shape of the equation <m>Ax=0</m>, we clearly have that the zero vector solves this equation, and so the solution set contains the origin (which solves the problem encountered in the last example). Now, suppose that <m>x</m> and <m>y</m> are solutions. Then <m>A(x+y)=Ax+Ay=0+0=0</m>, and so <m>x+y</m> also solves the system, and for a real number <m>\alpha</m>, <m>A(\alpha x)=\alpha Ax=\alpha0=0</m>, and so <m>\alpha x</m> is a solution too. Thus, the solution set satisfies parts 2 and 3 of the definition of a subspace, and hence is a subspace.
</li>
<li>
True. Suppose that <m>V</m> and <m>W</m> are both subspaces of <m>\mathbb R^n</m>. Then we can check each part of the definition is satisfied by the intersection of <m>V</m> and <m>W</m>. Firstly, since 0 is in both subspaces, its in the intersection, so it is a non-empty set. Furthermore, if <m>x</m> and <m>y</m> are in the intersection and <m>\alpha</m> is a real number, then <m>\alpha x+y</m> is in the intersection, using the axioms in the definition of subspace applied twice, once using the fact that <m>x</m> and <m>y</m> are in particular both in <m>V</m>, and once using that they are both in <m>W</m>. Note that this implies both the second and third parts of the definition of subspace, as we can take <m>\alpha=1</m> for part 2 of the definition, and choose <m>y=0</m> for part 3 of the definition.
</li>
<li>
False. Although the union of two intersections always contained the origin, it will usually not be a subspace. For instance, consider the two lines <m>y=\pm x</m> in <m>\mathbb R^2</m>. Their union is not closed under addition. To see this, we just need a counter example. For instance, choose <m>(1,-1)</m> on the line <m>y=-x</m> and the point <m>(2,2)</m> on the line <m>y=x</m>. In particular, both of these points are on the union of these two lines (which looks like an "X" in the plane). However, the sum of these two points, <m>(1,-1)+(2,2)=(3,-1)</m>, is not on the union of the two lines, as <m>3\neq\pm(-1)</m>.
</li>
</ol>
</li>
<li>
<ol>
<li>
Adding <m>-1/2</m> times the first equation to the second equation reduces us to the set of equations
<me>
\begin{aligned}
2x &+ &3y &= &3 \\
&- &(5/2)y &= &-15/2,
\end{aligned}
</me>
giving that the solution is <m>y=3</m> and <m>x=2</m>.
</li>
<li>
Here, we apply Gaussian elimination be adding <m>-3</m> times the first equation to the second equation, and adding the first equation to the second equation to give
<me>
\begin{aligned}
x & & &- &z &= &0 \\
& &y &+ &3z &= &1 \\
& &y & & &= &4
.
\end{aligned}
</me>
We then subtract the second equation from the third:
<me>
\begin{aligned}
x & & &- &z &= &0 \\
& &y &+ &3z &= &1 \\
& & & &-3z &= &3,
\end{aligned}
</me>
giving <m>x=-1</m>, <m>y=4</m>, and <m>z=-1</m>.
</li>
</ol>
</li>
<li>
If a system has a contradictory equation then it has no solution.
Otherwise, if there are any variables that are not leading a row
then it has infinitely many solution.
In the final case, where there is no contradictory equation and
every variable leads some row, it has a unique solution.
<ol>
<li>Unique solution</li>
<li>Infinitely many solutions</li>
<li> Infinitely many solutions</li>
<li> No solution</li>
<li>Infinitely many solutions</li>
<li>Infinitely many solutions</li>
<li>No solution</li>
<li>Infinitely many solutions</li>
<li>No solution</li>
<li>Unique solution</li>
</ol>
</li>
<li>
<ol>
<li>
We add <m>-1/2</m> times the first equation to the second to find
<me>\begin{aligned}
2x &+ &2y &= &5 \\
& &-5y &= &-5/2 ,
\end{aligned}</me>
which shows that <m> y=1/2 </m> and <m> x=2 </m> is the unique solution.
</li>
<li>
Gaussian elimination says that we should add the first and second equations:
<me>\begin{aligned}
-x &+ &y &= &1 \\
& &2y &= &3
\end{aligned}</me>
giving <m> y=3/2 </m> and <m> x=1/2 </m> as the only solution.
</li>
<li>
Row reduction (specifically, subtracting the first equation from the second)
<me>\begin{aligned}
x &- &3y &+ &z &= &1 \\
& &4y &+ &z &= &13
\end{aligned}</me>
shows, because the variable <m>z</m> does not have a pivot in the corresponding column, and hence is a free variable,
that there are many solutions.
</li>
<li>
Row reduction (subtracting 3 times the first equation from the second)
<me>\begin{aligned}
-x &- &y &= &1 \\
& &0 &= &-1
\end{aligned}</me>
shows that there is no solution.
</li>
<li>
Interchanging the first and fourth equations gives
<me>\begin{aligned}
x &+ &y &- &z &= &10 \\
2x &- &2y &+ &z &= &0 \\
x & & &+ &z &= &5 \\
& &4y &+ &z &= &20.
\end{aligned}</me>
Now subtracting the first equation from the third and adding <m>-2</m> times the first equation to the second equation yields
<me>\begin{aligned}
x &+ &y &- &z &= &10 \\
& &-4y &+ &3z &= &-20\\
& &-y &+ &2z &= &-5 \\
& &4y &+ &z &= &20 .
\end{aligned}</me>
Finally, we add the second and fourth equations and subtract <m>1/4</m> times the second row from the third row to obtain
<me>\begin{aligned}
x &+ &y &- &z &= &10 \\
& &-4y &+ &3z &= &-20\\
& & & &(5/4)z &= &0 \\
& & & &4z &= &0,
\end{aligned}</me>
which gives the unique solution<m> (x,y,z)=(5,5,0) </m>.
</li>
<li>
We begin by subtracting twice the first equation from the fourth equation and adding <m>-3/2</m> times the first equation to the third equation to obtain the system
<me>\begin{aligned}
2x & & &+ &z &+ &w &= &5 \\
& &y & & &- &w &= &-1 \\
& & &- &(5/2)z &- &(5/2)w &= &-15/2 \\
& &y & & &- &w &= &-1.
\end{aligned}</me>
Subtracting the second equation from the fourth gives us the system
<me>\begin{aligned}
2x & & &+ &z &+ &w &= &5 \\
& &y & & &- &w &= &-1 \\
& & &- &(5/2)z &- &(5/2)w &= &-15/2 \\
& & & & & &0 &= &0,
\end{aligned}</me>
which shows that there are many solutions.
</li>
</ol>
</li>
</ol>
</p>
</subsection>
</section>